Solve for x in R : 2 × sin(3x+4) + (√( 3 )) = 0


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Question Number 49765 by hassentimol last updated on 10/Dec/18

$Solve for x in R :$ $2 \times \sin (3 x + 4) + \sqrt{3} = 0$
Commented by mathmax by abdo last updated on 27/Feb/20

$(e) \Leftrightarrow s i n (3 x + 4) = - \frac{\sqrt{3}}{2} \Rightarrow s i n (3 x + 4) = s i n (- \frac{π}{3}) \Rightarrow$ $3 x + 4 = - \frac{π}{3} + 2 k π o r 3 x + 4 = \frac{4 π}{3} + 2 k π \Rightarrow 3 x = - 4 - \frac{π}{3} + 2 k π o r$ $3 x = - 4 + \frac{4 π}{3} + 2 k π \Rightarrow x = - \frac{4}{3} - \frac{π}{9} + \frac{2 k π}{3} o r x = - \frac{4}{3} - \frac{π}{9} + \frac{2 k π}{3}$
	Answered by afachri last updated on 10/Dec/18

	$2 \sin (3 x + 4) = - \sqrt{3^{}} \Rightarrow \sin (3 x + 4) = - \frac{\sqrt{3^{}}}{2}$ $consider 3 x + 4 = y, so : \sin (y) = - \sin (\frac{π}{3})$ $Sin has negative value in 3^{rd} and 4^{th} quadrant . p$ $∙ 3^{rd} quadrant \Rightarrow \sin (π + y) = - \sin (\frac{π}{3})$ $y = π + \frac{π}{3} = \frac{4 π}{3}$ $∙ 4^{th} quadrant \Rightarrow \sin (2 π - y) = - \sin (\frac{π}{3})$ $y = 2 π - \frac{π}{3} = \frac{5 π}{3}$ $\frac{4 π}{3} = 3 x + 4 \Rightarrow x_{1} = \frac{4 π - 12}{9}$ $\frac{5 π}{3} = 3 x + 4 \Rightarrow x_{2} = \frac{5 π - 12}{9}$
	Answered by mr W last updated on 10/Dec/18

	$\sin (3 x + 4) = - \frac{\sqrt{3}}{2}$ $\Rightarrow 3 x + 4 = 2 n π - \frac{π}{3} o r (2 n + 1) π + \frac{π}{3}$ $\Rightarrow x = \frac{(6 n - 1) π}{9} - \frac{4}{3} o r \frac{2 (3 n + 2) π}{9} - \frac{4}{3}$ $w i t h n = 0, \pm 1, \pm 2, . . .$
	Commented by afachri last updated on 10/Dec/18

	$did i solve wrong sir ? ? ? please let me know$ $Sir :) hope you {don}^{'} t mind$
	Commented by mr W last updated on 10/Dec/18

	$y o u r s o l u t i o n i s n o t c o m p l e t e s i r .$ $y o u t a k e o n l y 3 x + 4 = π + \frac{π}{3}, b u t i t c a n a l s o$ $b e 3 π + \frac{π}{3}, 5 π + \frac{π}{3}, e t c . - π + \frac{π}{3}, - 3 π + \frac{π}{3}, e t c .$ $y o u t a k e o n l y 3 x + 4 = 2 π - \frac{π}{3}, b u t i t c a n a l s o$ $b e 4 π - \frac{π}{3}, 6 π - \frac{π}{3}, e t c . - \frac{π}{3}, - 2 π + \frac{π}{3}, e t c .$ $a s t h e q u e s t i o n s a y s x \in R, i . e . - \infty < x < + \infty .$
	Commented by hassentimol last updated on 10/Dec/18

	$Sir, at the 2 nd line, {isn}^{'} t it 2 n π - \frac{2 π}{3} instead$ $for the second one . . . ?$
	Commented by afachri last updated on 10/Dec/18

	$thanks a lot for your correction and explanation$ $Mr . W, appreciate that . i^{'} ll fix my job again .$ $h ope you never be bored Sir . :) :)$
	Commented by mr W last updated on 10/Dec/18

	$2 n π - \frac{2 π}{3} i s t h e s a m e a s (2 n + 1) π + \frac{π}{3}$ $b e c a u s e n i s a n y i n t e g e r h e r e .$
	Commented by hassentimol last updated on 10/Dec/18

	$Ok . . . thanks sir I^{'} ve forgotten it . . .$
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